a) ( a - b + c ) - ( -b - a + c ) - [ - ( -a ) ]

= a - b + c + b + a - c - a

= 0

chắc là z ~~ 

a,=a+b-c-a-b

=(a-a)+(b-b)-c

=-c

a,(a+b)-(-c+a+b)
= a+b-c-a-b
= -c
b,-(x+y)+(-z+x+y)
=  -x-y+(-z)+x+y
=  -x+(-y)+x+y+(-z)
=-z
c,(m-n+p)+(-m+n+p)
= m-n+p+(-m)+n+p
=2p

Chứng minh đẳng thức sau:

a)(a-b)+(c-d)-(a+c)=-(b+d)

b)(a-b)-(c-d)+(b+c)=a+d

Bài 4: Đơn giản các biểu thức sau khi bỏ dấu ngoặc

a/ (a + b - c) - (b - c + d)

= a + b - c - b +c - d

= a + (b - b) + (-c + c) - d

= a - d

b/ -(a-b+c)+(a-b+d)

= -a + b - c + a - b + d

= (-a + a) + (b - b) - c + d

= -c + d

c/ (a+b)-(-a+b-c)

= a + b + a - b + c

= 2a + c

d/ -(a+b) + (a+b+c)

= -a - b + a + b + c

= c

xin lỗi\(^{100}\)

\(B=\left|x+10\right|+\left|y-10\right|+178\) ( \(x,y\) ∈ Z )

Ta có :

\(\left|x+10\right|\ge0\) ∀ \(x\)

\(\left|y-10\right|\ge0\) ∀ \(y\)

\(\Rightarrow\left|x+10\right|+\left|y-10\right|\ge0\)

\(\Rightarrow\left|x+10\right|+\left|y-10\right|+178\ge0+178\)

\(\Rightarrow B\ge178\) . Dấu '' = '' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+10=0\\y-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0-10=-10\left(TM\right)\\y=0+10=10\left(TM\right)\end{matrix}\right.\)

Vậy Min B = 178 \(\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=10\end{matrix}\right.\)

2.

A = 5 - \(\left|x-3\right|\)

Ta có :

\(-\left|x-3\right|\le0\) ∀ \(x\)

\(\Rightarrow5-\left|x-3\right|\le0+5\)

\(\Rightarrow A\le5\) . Dấu '' = '' xảy ra ⇔ x - 3 =0

⇔ x = 0 + 3 = 3

Vậy Max A = 5 ⇔ x = 3

\(A=\left(x+y+z\right)^3-\left(x+y-z\right)^3-\left(y+z-x\right)^3+\left(z+x-y\right)^3\)

Đặt \(B=\left(x+y+z\right)^3-\left(x+y-z\right)^3\)

\(=\left(x+y\right)^3+3z\left(x+y\right)^2+3\left(x+y\right)\cdot z^2+z^3-\left(x+y\right)^3+3z\left(x+y\right)^2-3\left(x+y\right)\cdot z^2+z^3\)

\(=6z\left(x+y\right)^2+2z^3\)

\(C=-\left(y+z-x\right)^3+\left(z+x-y\right)^3\)

\(=\left(x-y+z\right)^3+\left(x-y-z\right)^3\)

\(=\left(x-y\right)^3+3\left(x-y\right)^2\cdot z+3\left(x-y\right)\cdot z^2+z^3+\left(x-y\right)^3-3\left(x-y\right)^2\cdot z+3\left(x-y\right)\cdot z^2-z^3\)

\(=2\left(x-y\right)^3+6\left(x-y\right)\cdot z^2\)

=>\(A=6z\left(x+y\right)^2+2z^3+2\left(x-y\right)^3+6z^2\left(x-y\right)\)

a) Ta có: \(A=\left(-\dfrac{1}{3}x^2y^4\right)\cdot\left(-\dfrac{3}{5}x^3y\right)^2\)

\(=\dfrac{-1}{3}x^2y^4\cdot\dfrac{-9}{5}x^6y^2\)

\(=\left(\dfrac{-1}{3}\cdot\dfrac{-9}{5}\right)\cdot\left(x^2\cdot x^6\right)\cdot\left(y^4\cdot y^2\right)\)

\(=\dfrac{3}{5}x^8y^6\)

\(15\left(2a^2-1\right)+5\left(3-\frac{1}{5a}-6a^2\right)\)

\(=30a^2-15+15-\frac{1}{a}-30a^2\)

\(=-\frac{1}{a}\)

tại \(a=2017\)=> M= \(\frac{-1}{a}=\frac{-1}{2017}\)

\(\left(x-y\right)\left(x^2+xy+y^2\right)+y^3\)

\(=x^3-y^3+y^3\)

\(=x^3\)

ại \(x=2\)=> N= \(x^3=2^3=8\)